Steady State Response of a RL Circuit with Sinusoidal Input
The figure below shows the RL circuit from problem 10.9-6 of
Introduction to Electric Circuits, 5e by R.C. Dorf and J.A. Svoboda. The input to this circuit is the current
provided by the current source. (M is the magnitude of the input, in Amps, and is phase angle of the input, in degrees.) The response, or output, of the circuit is the current i2(t). The network function
describes the relationship between the phasors corresponding to the input, i1(t), and
the output, i2(t). In this case so
The steady state response of this circuit is given by
The values of the parameters M and of the input sinusoid, as well as the
values of the resistance R and inductance L, can be changed using the scrollbars.
- Set M=2 A, =-15 deg, R=8 Ohms and L = 2 H to see the circuit in Problem 10.6-9.
- Set M=5 A, R=8 Ohms and L = 2 H. Vary .
- Set M=5 A, =45 deg and L = 5 H. Vary R.
- Set M=5 A, =45 deg and R=10 Ohms. Vary L.
- Set M=5 A. Vary R and L until i2(t) is half as large as i1(t). Reconsider the equations given above for i1(t) and i2(t) in light of the values that you found for R and L.
- Set M=5 A. Vary R and L until the phase difference between i2(t) and i1(t) is 45 degrees. Reconsider the equations given above for i1(t) and i2(t) in light of the values that you found for R and L.